Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = 27{x^2}{e^{3x}}$$
\eqalign{ & y = \left( {9{x^2} - 6x + 2} \right){e^{3x}} \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\left( {9{x^2} - 6x + 2} \right){e^{3x}}} \right] \cr & {\text{use the product rule }} \cr & \frac{{dy}}{{dx}} = \left( {9{x^2} - 6x + 2} \right)\frac{d}{{dx}}\left[ {{e^{3x}}} \right] + {e^{3x}}\frac{d}{{dx}}\left[ {9{x^2} - 6x + 2} \right] \cr & {\text{use the formula }}\frac{d}{{dx}}{e^u} = {e^u}\frac{{du}}{{dx}}{\text{ for }}\frac{d}{{dx}}\left[ {{e^{3x}}} \right] \cr & \frac{{dy}}{{dx}} = \left( {9{x^2} - 6x + 2} \right)\left( {{e^{3x}}} \right)\frac{d}{{dx}}\left[ {3x} \right] + {e^{3x}}\frac{d}{{dx}}\left[ {9{x^2} - 6x + 2} \right] \cr & {\text{solve the derivatives using the power rule for differentiation}} \cr & \frac{{dy}}{{dx}} = \left( {9{x^2} - 6x + 2} \right)\left( {{e^{3x}}} \right)\left( 3 \right) + {e^{3x}}\left( {18x - 6} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = 3{e^{3x}}\left( {9{x^2} - 6x + 2} \right) + \left( {18x - 6} \right){e^{3x}} \cr & {\text{factoring}} \cr & \frac{{dy}}{{dx}} = \left[ {3\left( {9{x^2} - 6x + 2} \right) + \left( {18x - 6} \right)} \right]{e^{3x}} \cr & \frac{{dy}}{{dx}} = \left( {27{x^2} - 18x + 6 + 18x - 6} \right){e^{3x}} \cr & \frac{{dy}}{{dx}} = 27{x^2}{e^{3x}} \cr}